求∫根号下(2ax-x²)dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/20 10:46:56
x){Qgv>dG=FjʆtT$LΆhju J**@5
PDPHa&SdBbqf^NJ_\RdX
XFLi" e $Py}#M
Cm:#d#T 8( 1XTHјAQT
DjCM@aaXd~qAb(
1
求∫根号下(2ax-x²)dx
求∫根号下(2ax-x²)dx
求∫根号下(2ax-x²)dx
∫ √(2ax - x²) dx
= ∫ √[a² - (x - a)²] dx
令x - a = asinz,dx = acosz dz
= ∫ |acosz|(acosz) dz
= a²∫ cos²z dz
= (a²/2)∫ (1 + cos2z) dz
= (a²/2)z + (a²/2)sinzcosz + C
= (a²/2)arcsin[(x - a)/a] + (a²/2)[(x - a)/a][√(2ax - x²)/a] + C
= (a²/2)arcsin(x/a - 1) + (1/2)(x - a)√(2ax - x²) + C